Question

Show that the ꯳PQRS formed by P(2,1), Q(-1,3), R(-5,-3) and S(-2,-5) is a rectangle

Answer 1

$\bf{Given,}$
P ≡ (2,1) , Q ≡(-1, 3) , R ≡(-5,-3) and S ≡(-2,-5)
Now, first of all find side length PQ , QR , RS and SP
PQ = $\bf{\sqrt{(-1-2)^2+(3-1)^2}=\sqrt{9+4}=\sqrt{13}}$
QR = $\bf{\sqrt{(-5+1)^2+(-3-3)^2}=\sqrt{16+36}=\sqrt{52}}$
RS = $\bf{\sqrt{(-2+5)^2+(-5+3)^2}=\sqrt{9+4}=\sqrt{13}}$
SP = $\bf{\sqrt{(-2-2)^2+(-5-1)^2}=\sqrt{16+36}=\sqrt{52}}$

You observed , PQ = RS and QR = SP
It mean, PQRS must a parallelogram
[note :- you can find midpoint of diagonal AC and BD , you will be observe , midpoint of AC = midpoint of BD ]

More detail require , let's find PR,
PR = $\bf{\sqrt{(-5-2)^2+(-3-1)^2}=\sqrt{49+16}=\sqrt{65}}$
Here, you observed PQ, QR and PR follow Pythagoras theorem,
e.g., PQ² + QR² = 13 + 52 = 65 = PR²
So, ∠PQR = 90°

We find condition
PQ = RS and QR = SP
∠PQR = 90°
Hence, PQRS is a rectangle